## Continuous + holomorphic on a dense open => holomorphic?

Let D ⊂ ℂ be the closed unit disc in the complex plane, and let C be a continuously embedded path in D between the points -1 and 1. The curve C splits D into two halfs $D_1$ and $D_2$.

Let f : D→ℂ be a continuous function that is holomorphic on the interiors of $D_1$ and $D_2$.
Is f then necessarily holomorphic?

PS: If he path C is sufficiently smooth (so that ∫C f(z) dz makes sense), then f is necessarily holomorphic, as it is then given by Cauchy's formula $f(w)=1/(2\pi i)\int _ {\partial D} f(z) /(z-w) dz$.

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In the title, you assume that the complement of $C$ is dense in $D$, but in the body of the question you don't. Since $C$ could be a space filling curve, this isn't automatic. Do you want to assume that the complement of $C$ is dense? – David Speyer Aug 6 2010 at 13:31
@David: C cannot be space filling, since it is an embedded path, in particular homeomorphic to its image. It is also easy to see that the fact that C is embedded implies that its complement is dense, since C cannot contain balls. – Andrea Ferretti Aug 6 2010 at 13:42
An easy counterexample to the question in the title, as opposed to the body, of the question: Let $u:(0,1)\rightarrow (0,1)$ be the "Devil's Staircase" (Cantor function), let $D$ be the unit square $\{0\leq \mathrm{Re}(z),\mathrm{Im}(z)\leq 1\}\subset\mathbb{C}$, and let $f:D\rightarrow \mathbb{C}$ be the function $f(z)= u(\mathrm{Im}(z)) + iu(\mathrm{Im}(z))$. This is continuous and holomorphic in the dense open $U\subset D$ where the imaginary part of $z$ is not in the middle third Cantor set. Note that the complement of $U$ in $D$, while not a rectifiable simple closed curve, is measure 0. – Sam Lichtenstein Aug 7 2010 at 16:04
@Sam: Yes, I was aware of that counterexample. My title is misleading, since it asks another question than the one I care about. (apologies) – André Henriques Aug 7 2010 at 17:06
If $C$ contains points of the boundary circle other than $\pm 1$ then it doesn't necessarily split $D$ in two halves. Did you mean to exclude this possibility? – Victor Protsak Aug 7 2010 at 19:04
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Denjoy makes a detailed study of this question, and in particular constructs counterexamples where the curve C is the graph of a continuous function. Apparently, the construction works for curves which are 'very' non rectifiable, i.e., the local variation is infinite of a suitably high order at each point.

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If the curve C is rectifiable then the answer is yes.Under the assumption that C is rectifiable your question is known as Painleve's Theorem.It follows from the strong version of Cauchy's theorem which is stated as follows: If C is a simple closed rectifiable curve in the plane and f is holomorphic in the interior and continuous in the closed bdd region enclosed by C then the integral of f over C is zero.See for example the book by Behnke and Sommer page 119 (the book is in German).You can also find a proof of the strong form of Cauchy's theorem in the book titled:Elements of the topology of plane sets of points by M H A Newman,2nd edn page 187. If the jordan arc has positive area the answer to the question is no.See pages 122- 123 of the article in the amer math monthly vol 81 no 2 pages 115-137 year 1974.The paper is by Lawrence Zalcman who has other papers on this topic.

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"Rectifiable" implies "Lebegue measure zero". So Andrea Ferreti's answer seems to cover a wider class of curves. Right? – André Henriques Aug 7 2010 at 17:11
Certainly, but I have some difficulty understanding the continuity part of the argument. – Mohan Ramachandran Aug 7 2010 at 17:30
But C completely divides D in two halves! I believe your argument if C is entirely inside D (meaning that an open neighbourhood of C is in D). Isn't Painleve's theorem is about extending Riemann maps to the boundary in a differentiable manner? Can you please elaborate a bit? – timur Nov 2 2010 at 20:01

If the image of the curve has measure 0, it seems true. Indeed it is enough to prove that both the real and imaginary part are harmonic. This will be true if they satisfy the mean value identity on small balls.

The mean value identity is clear outside of $C$; on points of $C$ it follows by continuity and the fact that the integral of $f$ on $C$ with respect to the $2$-dimensional Lebesgue measure is $0$.

EDIT: I'm sorry, according to Mohan Ramachandran comment below, this answer is wrong.

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I don't understand the "follows by continuity" part of the argument. Could you please elaborate? – André Henriques Aug 6 2010 at 21:02
your argument seems to imply analytic extension across any compact subset of the open unit disc of measure zero.this is not true see Urysohn Fund.Math vol 4 pages 144-150 – Mohan Ramachandran Aug 7 2010 at 18:34
Mean value property will not follow since the values of the function on upper part of the domain do not "know" what the values on the lower part is doing. To connect these values you need something like the Schwarz reflection principle. By the way, instead of the mean value property Morera's theorem gives a direct approach. – timur Nov 2 2010 at 19:53
I'm not sure, but maybe in this answer the integral of $f$ along the path $C$ is confused with the Lebesgue integral of $f$ on the image of $C$ ? – Qfwfq Feb 27 2011 at 16:51
No, it is not. The problem is to verify the mean value identity on small balls centered on points of $C$, which is not as trivial as I envisaged. A simple continuity argument does not work, as I found by trying to add more details. – Andrea Ferretti Feb 27 2011 at 17:29

We can go via Morrera's Theorem as well:

Let T be a triangle in the disc: partition the preimage of your curve $\cap T$ into n pieces (in a suitably 'going to be dense when $n \to \infty$' sort of way), now barycentrically subdivide T until we have a simplicial approximation of the curve. Since $f$ is continuous (and so has no singularities) and C has measure zero, as $n \to \infty$ the intergral around the subtriangles containing the curve must tend to $0$, the remainder are zero by Cauchy.

In steps Morrera, and the day is saved.

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The point is that I do not see how we can be sure that C has measure 0. Maybe I'm just missing some trivial thing. – Andrea Ferretti Aug 6 2010 at 15:13
If C is homeomorphic to its preimage, it necessarily has dimension 1 and so L^2 measure zero. – Tom Boardman Aug 6 2010 at 15:27
For the life of me though, I can't find a reference for that last implication other than wikipedia... :( Anyone have a measure theory text to hand? – Tom Boardman Aug 6 2010 at 15:45
There are Jordan arcs of positive area.These were first constructed by W F Osgood. – Mohan Ramachandran Aug 6 2010 at 17:07
Damn you wikipedia!!! en.wikipedia.org/wiki/… -I thought I had seen it somewhere else as well though... Although in hindsight, that may have just been wikipedia a while back... – Tom Boardman Aug 6 2010 at 22:32

Not an answer, but too big for a comment and useful for many similar problems.

From Chapter VI of Theory of the Integral by Stanislaw Saks, p.197, available online at http://banach.univ.gda.pl/pdf/saks/

For a domain $G$:

Theorem (Attributed to Besicovitch) Let $f$ be continuous on $G$, satisfying:

(i): $f'(z)$ exists for each $z \in G \setminus E_1$,

(ii): $$\limsup_{h \to 0} \left| \frac{f(z+h) - f(z)}{h} \right| < \infty$$
for all $z \in G \setminus E_2$.

(iii): $E_1$ has Lebesgue measure zero.

(iv): $E_2$ is a countable union of finite-length curves.

Then: $f$ is holomorphic on $G$.

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